Use de Moivre's theorem to prove that
$$\tan 3 \theta \equiv \frac { \tan \theta \left( 3 - \tan ^ { 2 } \theta \right) } { 1 - 3 \tan ^ { 2 } \theta } .$$
(a) By putting \(\theta = \frac { 1 } { 12 } \pi\) in the identity in part (i), show that \(\tan \frac { 1 } { 12 } \pi\) is a solution of the equation
$$t ^ { 3 } - 3 t ^ { 2 } - 3 t + 1 = 0 .$$
(b) Hence show that \(\tan \frac { 1 } { 12 } \pi = 2 - \sqrt { 3 }\).
Use the substitution \(t = \tan \theta\) to show that
$$\int _ { 0 } ^ { 2 - \sqrt { 3 } } \frac { t \left( 3 - t ^ { 2 } \right) } { \left( 1 - 3 t ^ { 2 } \right) \left( 1 + t ^ { 2 } \right) } \mathrm { d } t = a \ln b$$
where \(a\) and \(b\) are positive constants to be determined.